[VAR]=Notes on variational calculus

As above one can make precise this condition on u by
d
dε J[u + εη] ∣
∣∣∣ε=0
= 0 (26)
for all functions η = η(x, y) on Ω. The following computation similar to the one in
(11),
∣
d ∣∣∣ε=0
dε J[u + εη]
= d ∫
F(u(x, y) + εη(x, y), u x (x, y) + εη x (x, y), u y (x, y) + εη y (x, y), x, y)dxdy
dε Ω
∫ ( ∂F
∂F
= η(x, y) +
Ω ∂u(x, y) ∂u x (x, y) η x(x, y) + ∂F )
∂u
} {{ }
y (x, y) η y(x, y) dxdy
} {{ }
d
dx ( ∂F
∂ux(x,y) η(x,y))−η(x,y) d ∂F
···
dx ∂ux(x,y)
∫ ( ∂F
= BT + η(x, y)
∂u(x, y) − d ∂F
dx ∂u x (x, y) − d ∂F
)
dxdy (27)
dy ∂u y (x, y)
Ω
where
∫
BT =
Ω
( d
dx ( ∂F d
η(x, y)) +
∂u x (x, y) dy ( ∂F
)
∂u y (x, y) η(x, y)) dxdy
∮ (
∂F
∂F
)
= η(x, y)dy −
∂u x (x, y) ∂u y (x, y) η(x, y)dx
∂Ω
(28)
is a boundary term (we used Green’s theorem) which vanishes since u is fixed on ∂Ω,
and thus the allowed variation functions η vanish on ∂Ω. We thus can conclude, as
above:
Fact E: All solutions of the Problem E above satisfy the following Euler-Lagrange
equations
∂F
∂u(x, y) − d ∂F
dx ∂u x (x, y) − d ∂F
dy ∂u y (x, y) = 0 (29)
where u| ∂Ω is fixed.
Again, it is imporant to distinguish partial and total derivatives, e.g.,
d ∂F
= ∂2 F
u x + ∂2 F
u xx +
∂2 F
u xy + ∂2 F
dx ∂u x ∂u∂u x ∂u y ∂u x ∂x∂u x
∂u 2 x
(we suppress arguments x, y of u and its derivatives) etc.
Similarly as above there are also modified variational problems where u is allowed
to vary on all of ∂Ω or on parts of it. In this case one also gets the Euler-Lagrange
12